Question

A survey asked whether respondents favored or opposed the death penalty for people convicted of murder. Software shows the results below, where X refers to the number of the respondents who were in favor. Construct the 95% confidence interval for the proportion of the adults who were opposed to the death penalty from the confidence interval stated below for the proportion in favor. (Round to three decimal places as needed.)
X = 1,790

N = 2,610

Sample p = 0.6858

95% Confidence Interval = (0.668, 0.704)

Answer 1

Answer:

95% confidence interval for the proportion of the adults who were opposed to the death penalty is (0.668, 0.704).

Step-by-step explanation:

We are given that a survey asked whether respondents favored or opposed the death penalty for people convicted of murder. Software shows the results below, where X refers to the number of the respondents who were in favor.

X = 1,790

N = 2,610

$\hat p$ = Sample proportion = X/N = 0.6858

Firstly, the pivotal quantity for 95% confidence interval for the population proportion  is given by;

     P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion = 0.6858

           n = sample of respondents = 2,610

           p = population proportion

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population​ proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at

                                   2.5% level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$]

  = [ $0.6858-1.96 \times {\sqrt{\frac{0.6858(1-0.6858)}{2610} }$ , $0.6858+1.96 \times {\sqrt{\frac{0.6858(1-0.6858)}{2610} }$ ]

  = [0.668 , 0.704]

Therefore, 95% confidence interval for the population proportion of the adults is (0.668, 0.704).